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4sin²(x)-1=0=(2sin(x)-1)(2sin(x)+1), so sin(x)=½ or -½ and x=π/6 or -π/6, where n is any integer.

Since sine is a cyclic function we can write a general solution x=πn±π/6. This gives a range of values for x:

π/6 (30°), 5π/6 (150°), 7π/6 (210°), 11π/6 (330°), 13π/6 (390°), ... as well as all corresponding negatives.

by Top Rated User (660k points)

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